3-Env. Engg.-II Lab Civ-10-36

7/27/2019 3-Env. Engg.-II Lab Civ-10-36

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2010-CIV-36 [SAFEER ULLAH KHAN]

Environmental Engineering-II (Lab Manual) Page 1

DISSOLVED OXYGEN

TITLE

To determine the amount of Dissolved Oxygen (DO) in water.

DISSOLVED OXYGENThe dissolved oxygen (DO) is oxygen that is dissolved in water. The oxygen dissolves by

diffusion from the surrounding air; aeration of water that has tumbled over falls andrapids; and as a waste product of photosynthesis. A simplified formula is given below:

Photosynthesis (in the presence of light and chlorophyll):

Carbon dioxide + Water --------------> Oxygen + Carbon-rich foods

CO2 H2O O2 C6H12O6

Fish and aquatic animals cannot split oxygen from water (H2O) or other oxygen-

containing compounds. Only green plants and some bacteria can do that throughphotosynthesis and similar processes. Virtually all the oxygen we breathe ismanufactured by green plants. A total of three-fourths of the earths oxygen supply is

produced by phytoplankton in the oceans.

Fish, invertebrates, plants, and aerobic bacteria all require oxygen for respiration. Much of the dissolved oxygen in water comes from the atmosphere. After dissolving

at the surface, oxygen is distributed by current and turbulence. Algae and rooted

aquatic plants also deliver oxygen to water through photosynthesis. The main factor contributing to changes in dissolved oxygen levels is the build-up of

organic wastes. Decay of organic wastes consumes oxygen and is often concentrated

in summer, when aquatic animals require more oxygen to support highermetabolisms.

Depletions in dissolved oxygen can cause major shifts in the kinds of aquaticorganisms found in water bodies.

Temperature, pressure, and salinity affect the dissolved oxygen capacity of water. Theratio of the dissolved oxygen content (ppm) to the potential capacity (ppm) gives thepercent saturation, which is an indicator of water quality.

METHODS OF DETERMINATION OF DO IN WATER:

(1)

Winkler Method(2) Azide modification of Winkler Method(3) Rideal Stewart method(4) DO-meters
http://www.lenntech.com/Periodic-chart-elements/O-en.htmhttp://www.lenntech.com/Periodic-chart-elements/O-en.htm


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(1) Winkler MethodIts principle is that oxygen oxidizes manganese (Mn

+2) to higher oxidation state then this

higher oxidation state manganese converts iodide ion (I-1

) to iodine I2 and amount of freeiodide librated is equivalent to DO

Mn

+2

+ 2OH

-1

Mn(OH)2Mn(OH)2 + 1/2O2

MnO2 + H2OMnO2 + 2I

-1+ 4H

+1Mn

+2+ I2 + H2O

Then by titration with S2O3-2

, thiosulphate ion (thiosulphate ion comes from sodiumthiosulphate), iodine can be calculated.

2S2O3-1

+ I2 S4O6-2

+ 2I-1

If we use 0.025 N Na2S2O3 then 1ml of titrant = 1mg of DO

(2) Azide modification of Winkler MethodIf nitrites NO2-1

will present in water, they will change the results because they convert

iodide ions to iodine before performance of experiment. To remove NO2-1

sodium azides

are added.

PROCEDURE

1. Take BOD bottle (300 ml volume)2. Fill the bottle with water sample.3. Add 1 ml MnSO4 solution to it and mix uniformly with the help of pipette.4. Add 1ml alkali azide iodide solution. On addition, if white ppts. are formed, then

there is no DO in water. Formation of Reddish brown ppts. Indicates the presenceof DO.

5. If Reddish brown ppts are formed, stopper the bottle and shake it upside down for20 times and allow the ppts. To settle down for about 2 inches.

6. Add 1 ml Concentrated H2SO4 and again shake for about 8 times.7. Take 200 ml of this water sample in a titration flask and titrate it with 0.025 N

Na2S2O3 till the appearance of light yellow color.8. Then add 1ml starch solution. The color of solution becomes blue on this

addition.

9. Again titrate it with 0.025N Na2S2O3 till the disappearance of blue color.10. Note the volume of titrant used.

ml of titrant used = DO in mg/liter

The above formula is applicable if we use 200ml of water sample solution and 0.025NNa2S2O3. General formula is given by

mlinsampleofx volF

8000xNxusedtitrantofmean vol(mg/l)DO


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Where,bottleBODofvol

usedreagentsofvol-bottleBODofvolF

Dissolved Oxygen (DO)

Observations And Calculations

Sampleno.

Sample

description

sample

volume Volume

of titrant

(Na2S2O3)

Dissolved

Oxygen

(DO)

Mean

Dissolved

oxxygen

(DO)

(ml) (ml) (mg/l) (mg/l)

1Lab

prepared

200 6.7 6.7676.8

100 3.4 6.868

2 Lab

prepared

200 6.2 6.266.06

100 2.9 5.85

3Lab

preparedNo Dissloved Oxygen

Questions

Q.1. What is importance of PALE YALLOW colour in titration for estimation of

dissolved oxygen?

The Starch-Iodine complex is not very much soluble in water, so the starch is

added nearthe end point of an Iodine titration, when the iodine concentration is

low. This eliminates error due the fact that some iodine remains absorbed on thecomplex and goes undetected. we add starch only near the end point when the

color of the solution is pale yellow so that the starch should not complex and go

lumpy.This also tells us whether the experiment is proceeding in the right direction. Like

if starch is added before this pale yellow, resulting color would be black. And if it

is added after that it would go colorless.

Q.2. Give applications of dissolved oxygen in Civil Engineering.

This test can help in making a decision that whether the water coming out fromthe treatment plant is good for drinking or the intend use or not. Which is itself is

an engineering application of it.

The water that is good for drinking is good for construction as well. Hence,Meanwhile clan water will help cement developing a good bond.


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Q.3. what are various sources of oxygen by which it can dissolve in water and

wastewater?

Sources of oxygen are: Atmosphere: A great part of oxygen that is present in water comes from Atmosphere.

After dissolving at the surface, oxygen is distributed by current and turbulence. Algae and rooted plants: they also are a cause of DO mainly due to the photosynthesis. Aeration: Water falls and similar situations may provide a chance of aeration when

contact area of water and air considerably increased and oxygen has more chances to

dissolve itself in the water.

Q.4. What will be DO of solution if 10mL of 0.2N sodium thiosulphate solution is

used in titration? (F=0.99)

Given Data:

D.O = ? N = 0.2 of Na2S2O3A = Volume of Titrant used = 10 ml Volume of Sample = 300 ml

Solution:

= 53.87 mg/l

Q.5. What will be DO of solution if 6.3mL, 6.2mL and 6.3mL of 0.2N sodium

thiosulphate solution is used in each of 100mL titration of solution ? (F=0.99)

Given Data:

D.O = ? N = 0.2 of Na2S2O3A = Volume of Titrant used = (6.3+6.2+6.3) ml = 18.8 ml

Volume of Sample = (100+100+100) ml = 300 ml

Solution:

= 101.28 mg/l

COMMENTS

DO of sample 1 is more than of sample 2 which makes it better foraquatic life and photosynthesis in green plants under water.

Sample 3 contained no dissolved oxygen.


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BIOCHEMICAL OXYGEN DEMAND (BOD)

TITLE

To determine the amount of Biochemical Oxygen Demand (BOD) in domestic wastewater.

Biochemical Oxygen Demand (BOD)

The amount of oxygen required by the bacteria while stabilizing decomposable organic

matter under aerobic conditions. Decomposable means that organic matter can serve as

food for the bacteria and energy is derived from its oxidation.

Biochemical oxygen demand is a measure of the quantity of oxygen used bymicroorganisms (e.g., aerobic bacteria) in the oxidation of organic matter.

Natural sources of organic matter include plant decay and leaf fall. However,plant growth and decay may be unnaturally accelerated when nutrients andsunlight are overly abundant due to human influence.

Urban runoff carries pet wastes from streets and sidewalks; nutrients from lawnfertilizers; leaves, grass clippings, and paper from residential areas, which

increase oxygen demand.

Oxygen consumed in the decomposition process robs other aquatic organisms ofthe oxygen they need to live. Organisms that are more tolerant of lower dissolved

oxygen levels may replace a diversity of more sensitive organisms.

BOD Level (in ppm) Water Quality

1 - 2 Very Good-not much organic waste present

3 - 5 Moderately clean

6 - 9 Somewhat polluted

10+ Very polluted

Importance of BOD Test in Environmental Engineering

The BOD test is used to determine the relative oxygen requirements of wastewaters,

effluents, and polluted waters. The test measures the oxygen utilized during a specified

incubation period for the biochemical degradation of organic material. It is also used todetermine treatment plant efficiency.

Determination of BODPrinciple

The method consists of filling with sample, to overflowing, an airtight bottle of thespecified size and incubating it at the specified temperature for 5 days. Dissolved oxygen


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is measured initially and after incubation, and the BOD is computed from the difference

between initial and final DO. Because the initial DO is determined shortly after the

dilution is made, all oxygen uptake occurring after this measurement is included in theBOD measurement.

Sampling and Storage

Sample for BOD analysis may degrade significantly during storage between collectionand analysis, resulting in low BOD values. Minimize reduction of BOD by analyzing

sample promptly or by cooling it to near-freezing temperature during storage. However,

even at low temperature, keep holding time to a minimum. Warm chilled samples to 20 3C before analysis.

Apparatus

Incubation bottles: Use glass bottles having 60 mL or greater capacity (300mLbottles having ground-glass stopper and a flared mouth are preferred).

Air incubator or water bath, thermo-statistically controlled at 20 1C. Excludeall light to prevent possibility of photosynthetic production of DO.

Reagents

Prepare reagents in advance but discard if there is any sign of precipitation or biological

growth in the stock bottles.

a. Phosphate buffer solution: Dissolve 8.5 g KH2PO

4, 21.75 g K

2HPO

4, 33.4 g

Na2HPO4.7H2O, and 1.7 g NH4CI in about 500 mL distilled water and dilute to 1

Lit. The pH should be 7.2 without further adjustment. Alternatively, dissolve 42.5g KH2PO4 or 54.3 g K2HPO4 in about 700 mL distilled water. Adjust pH to 7.2

with 30% NaOH and dilute to I Lit.

b. Magnesium sulfate solution: Dissolve 22.5 g MgS04.7H20 in distilled water anddilute to 1 L.

c. Calcium chloride solution: Dissolve 27.5 CaCl2 in distilled water and dilute to 1L.

d. Ferric Chloride solution: Dissolve 0.25 g FeCl3.6H2O in distilled water and diluteto 1 L.

e. Acid and alkali solution, 1N, for neutralization of caustic or acidic waste samples.1) Acid-Slowly and while stirring, add 28 mL cone. Sulfuric acid to distilled

Water. Dilute to 1 L. 2) Alkali-Dissolve 40 g sodium hydroxide in distilled water.Dilute to 1 L.

f. Sodium sulfate solution: Dissolve 1.575 g Na2SO3 in 1000 mL distilled water.This solution is not stable; prepare daily.

g. Nitrification inhibitor: 2-chloro-6-(trichloromethyl) pyridine (if nitrificationinhibition desired).


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h. Glucose-glutamic acid solution: Dry reagent-grade glucose and reagent-gradeglutamic acid at 103C for 1 h. Add 150 mg glucose and 150 mg glutamic acid to

distilled water and dilute to 1 L. Prepare fresh immediately before use.i. Ammonium chloride solution: Dissolve 1.15 g NH4CI in about 500 mL distilled

water, adjust pH to 7.2 with NaOH solution and dilute to 1 L. Solution contains

0.3 mg N mL

-1

j. Dilution water: Use demineralized, distilled, tap, or natural water for makingsample dilutions.

Procedure: (with out seeding)

1) First of all it is important to know the amount of samples to be used for test. Forthis purpose the source of sample is to be recorded which will indicate the

approximate value of BOD5 for the sample.(i) Domestic sewage BOD5 =100-500mg/L(ii) Effluent from treatment plant= 20-80mg/L(iii)

River water = 2-4mg/L2) Take 9 BOD bottles note their numbers and arrange them in 3 groups.

3) Fill each bottle half with dilution media ensuring that no air gets mixed with themedia while fill in as in DO test.

4) Add 2ml sample in each of the three bottles marked as first group; 5 ml in eachbottle of 2

ndgroup and 10ml in each bottle of the 3

rdgroup.

5) Fill the bottle completely with dilution media and place the stopper such that noair bubbles are trapped.

6) Now take one bottle from each set and estimate its DO. This will be DO initial orDO 0days.

7) For comparison prepare two more bottles with blank dilutions media (with outsewage sample) and find the DO from one bottle.

8) Place the rest of the six bottles with sewage samples and one bottle for blank inthe incubator at 20

0C.

9) After 5 days find out DO in all bottles.10)That value of oxygen depletion should be considered correct which gives an

oxygen depletion of at least 2 mg/L. and which have at least 0.5 mg/L DO after 5

days of incubation.11)Calculate BOD5 at 200C. for the sample using following relationship

.

BOD(mg/L) = DO depletion (mg/L) *300

Volume of sample in bottle (ml)


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BOD5


AT ZERO

DAYS

bottleno.

sampleadde

d

volumeof

sample Volume

of

Na2S2O3

Dissolved

oxygen

Mean

Dissolved

oxygen

(ml) (ml) (ml) (mg/l) (mg/l)

315 2 300 9.2 6.195

6.87

135 5 300 10.7 7.205

418 10 300 10.7 7.205

BLANK

AT FIVE

DAYS

bottle

no.

sam

ple

add

ed

volu

me

of

sam

le Volume

ofNa2S2O3

Dissolved

oxygen

Mean

Dissolvedoxygen

(ml) (ml) (ml) (mg/l) (mg/l)

12 2 300 8.3 5.59

5.695

10 2 300 8.6 5.8

92 5 300 7.1 4.78

5.52

25 5 300 9.3 6.26

126 10 300 8 5.38

4.98

70 10 300 6.8 4.58

BLANK


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DO DEPLETION

Bottle no.sample

added

DO at zero

days

DO at 5

days

DO

DepletedBOD5

(ml) (mg/l) (mg/l) (mg/l) (mg/l)

315, 12, 10 2 6.195 5.695 0.5 75

135, 92, 25 5 7.205 5.52 1.685 101.1

418,126,70 10 7.205 4.98 2.225 66.75

BLANK

MEAN BOD5 = 80.95 mg/l

QuestionsQ.1. What is the role of chemicals added in dilution media?

Bacteria derive energy for growth and other metabolites utilizing variety of

carbon and nitrogenous sources through oxidation and fermentation. Studies of such

activities are possible with bio-chemical media only carbon and nitrogen as supplementedas exclusive nutrient sources to basal medium (like peptane) e.g. sugar at 0.5-1% leve;s

are added to peptone water along with an indicator which imparts different color to a

medium at acidic and alkaline pH.

Q.2.Why aeration is important for dilution media?

Bacteria produce large quantities of acetic acid if the growth medium containslittle or no oxygen causing the growth medium to reach pH 4 and lower. At this pH,

bacteria growth slows down or even stopps. Thats why aeration is important for dilution

media.

Q.3. Give any FOUR differences between COD and BOD.

COD is the total measurement of all chemicals in the water that can be oxidized

while BOD is supported to measure the amount of food or organic carbons that bacteria

can oxidized.

Permissible limit of COD is 250-500 ppm but for BOD is 30 mg/l. COD is always greater than BOD. COD is not not different b/w bi-logically available and inert organic

matter and it measure total quantity of oxygen required to oxidize all

organic matter into CO2 and H2O.


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Q.4. What will be DO depletion after five days, if 12.1mL and 6.3mL of sodium

thiosulphate (0.025N) were used at zero day and at five day, respectively?

Volume of sample is 3mL.F= 0.99

Sample Added Volume ofsample Volume ofNa2S2O3Dissolvedoxygen DO Depletion

3 300 12.1

At zero days

8.148

3.90

3 300 6.3

At five days

4.242

Q.5. What will be BOD after five days, if 12.1mL and 6.3mL of sodium

thiosulphate (0.025N) were used at zero day and at five day, respectively?

(Volume of sample is 3mL &F= 0.99)

Sample Added Volume of

sample

Volume of

Na2S2O3

Dissolved

oxygen

DO Depletion BOD

(mg/l)

3 300 12.1

At zero days

8.148

3.90 390

3 300 6.3

At five days

4.242

COMMENTS

Value of DO after 5 days decreases because some oxygen is used byorganic matter and micro organisms.

BOD5 is greater than 10 ppm so sample is highly polluted.BOD5 for effluent of treatment plant must be less than 80, so if the

sample is a effluent of treatment plant, results are OK.


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CHEMICAL OXYGEN DEMAND (COD)

TITLE

Determination of amount of Chemical Oxygen Demand (COD) in water.

Chemical Oxygen Demand (COD)

The Chemical Oxygen Demand, or COD,is a measurement of the amount of material thatcan be oxidized (combined with oxygen) in the presence of a strong chemical oxidizing

agent. Since the COD test can be performed rapidly, it is often used as a rough

approximation of the water's BOD, even though the COD test measures some additional

organic matter (such as cellulose) which is not normally oxidized by biological action. Aswith the BOD test, the COD test is reported as mg/Lit of oxygen used. There is a specific

normal range of COD found in various kinds of domestic wastewater. Keep in mind that

the addition of industrial waste can cause these values to vary widely. Biochemical

oxygen demand is a measure of the quantity of oxygen used by microorganisms (e.g.aerobic bacteria) in the oxidation of organic matter.

METHODS OF DETERMINATION OF COD1. Open Reflux Titrimetric Method

Principle

In this method known amount of strong oxidizing agent is being added. Then reaction

takes place to form CO2 and H2O. Then remaining amount of oxidizing agent is being

determined by titration. The amount of oxidizing agent to be added depends upon theCOD of sample which can roughly be known by knowing the source of sample.

Equipment:

Caution:

The presence of minute traces of organic matter on the equipment will cause large errorsin the test results. So clean all equipment thoroughly before using.

Erlenmeyer flask Small beaker Titration apparatus:

25 or 50 mL burette, graduated in 0.1 mL burette support 100 mL graduated cylinder rubber-tipped stirring rod, or magnetic stirrer and stir bar white porcelain evaporating dish, 4.5 inches in diameter

Reflux apparatus:


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500 or 250 mL Erlenmeyer flasks with ground glass 24/40 neck300 mm jacket Liebig, West, or equivalent condenser with 24/40 ground-glass

jointhot plate with sufficient power to produce at least 1.4 W /cm2 of heating surface

Blender Pipets Glass beads Fume hoodReagents

Standard potassium dichromate solution, 0.25N or O.025N Sulfuric acid reagent containing silver sulfate catalyst Standard ferrous ammonium sulfate titrant Ferroin indicator solution Mercuric sulfate crystals Sulfamic acid Concentrated sulfuric acid Distilled waterTheory of Titration

The COD analysis, by the dichromate method, is more commonly used to control and

continuously monitor wastewater treatment systems, The COD of an effluent is usually

higher than the BOD5 since the number of compounds that can be chemically oxidized is

greater than those that can be degraded biologically, It is also common to make a

correlation of BOD5 versus COD and then use the analysis of COD as a rapid means ofestimating the BOD5 of a wastewater. This may be convenient since only about three

hours are needed for a COD determination while a BOD5 takes at least 5 days. However,this procedure can be used only for specific situations where there is low variability in

the composition of a wastewater, and the results of a system cannot be used reliably in

other cases.

The method of COD which uses dichromate as oxidant is carried out by heating under

total reflux a wastewater sample of known volume in an excess of potassium dichromate


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(K2Cr2O7) in presence of sulphuric Acid (H2SO4) for a fixed period (usually two hours)

in presence of silver sulphate (Ag2SO4) as catalyst. The organic matter present is

oxidized and, as a result, the dichromate ion (orange colour) is consumed and replacedby the chromic ion (green colour):

Cr2O7-2

+ 14H+

+ 6e-

2Cr3+

+ 7H2O

The COD is calculated by titrating the excess of dichromate or by spectrophotometrically

measuring the Cr+3

ions at 606 nm. Another possibility is to measure the excessdichromate spectrophotometrically at 440 nm. Titration requires more work but is

considered more precise.

The presence of silver sulphate as catalyst is needed for complete oxidation of aliphaticcarbon compounds. The standard method implies cooling of the sample after the two hour

digestion period, adding a few drops of indicator (ferroin) solution and titrating the

excess dichromate with a solution of ferrous ammonium sulphate of knownconcentration, until the colour changes from brilliant green to reddish brown. The

titration reaction corresponds to the oxidation of the ferrous ammonium sulphate by the

dichromate:

Cr2O7-2

+ 14H+

+ 6Fe+2

2Cr3+

+ 6Fe+3

+ 7H2O

The change in colour corresponds to the formation of the complex ferrous ion

phenanthroline which occurs when all the dichromate ion has been reduced to Cr3+

.

(Fe(C12H8N2)3)3+

+ e (Fe(C12H8N2)3)2+

Ferric Phenanthroline Ferrous Phenanthroline

(Green Blue) (Reddish Brown)

Interferences

A common interference factor in the COD test is the presence of chlorides. If seawater isused at some point in the processing or salt brines are used for some "curing" operations,

chlorides will most probably appear in the wastewater causing interference while they are

oxidized by the dichromate:

Cl-+ Cr2O7

2-+ 14H

+3Cl2 + 2Cr

3++ 7H2O

This interference causes erroneously high values of COD which can be prevented by theaddition of mercuric sulphate (HgSO4) which reacts to form mercuric chloride and

precipitates:

Hg2+ + 2Cl- HgCl2


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Procedure:

1) Place 50ml sample in 500ml refluxing flask (for samples with COD>900mg/L usea smaller sample diluted to 50ml).

2) Add 1g HgSO4 and several glass beeds.3) Add slowly 5ml H2SO4 reagent while mixing to dissolve HgSO44) Cool while mixing to avoid the loss of volatile materials.5) Add 25 ml 0.25N K2 Cr2O7 solution and mix.6) Attach the flask to the condenser and turn on cooling water.7) Add remaing H2SO4 (70ml) through open end of the condenser continue mixing

while adding H2SO4.

8) Reflux the mixture for 2 hrs and cool to room temperature, after diluting themixture to about twice its volume with distilled water.

9) Titrate excess of K2 Cr2O7 with Ferrous ammonium sulfate using 2,3 drops offerrion indicator. The end point will be from blue green to reddish brown.

10)Reflux and titrate in the same manner a blank containing the reagents and thevoume of the distilled water will be equal to that of sample.

COD


COD (mg/l) = (A-B)xNx8000/(Vol. of sample, ml)N=0.25

Volume of sample = 50 ml

Serialno.

Descriptionof

Sample

volumeof

titrantusedfor

sample

Volume of

Titrant

used for

Blank

Chemical

Oxygen

Demand

COD

B A (mg/l)

1 Sample-1 8.5 16.1 304

2 Sample-2 12.5 18 220


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Questions

1. Calculate COD of an industrial solution if 5mL sample is used against FAS.The volume of FAS (0.25N) used against sample is 12.34mL while for blank it

is 23.1mL.

A = 23.10 ml

B = 12.34 ml

N = 0.25 N

= 4304 mg/l

2.

What will be the volume of FAS for sample if COD of sample is 424.5mg/L ifFAS volume for blank is 30.1mL? Use standard procedure for calculations.

(N=0.25)

COD = 424.5 mg/l

A = 30.1 ml

B = ?Vol. of sample = 50 ml

N = 0.25 N

B = 19.487 ml

3. Chloride ions produce interference in COD measurements. It can beremoved either by silver or mercury salts. Why only these two salts are used?

Can we use any other salt?

The most common interference is the chloride ions. Chlorides react with silver

ions to make ppt. silver chloride and thus inhibit the catalytic activity of silver.Bromides, iodides and any other reagents that inactivates the silver ions can infer

similarly.

The interfernces are negative in that they tend to restrict the oxidizing action of

dichromate ions itself. However, under the rigorous digestion procedures foranalysis, Cl

-,Br

-,I

-can react with dichromate to produce different chemical forms

of halogens and the chromic ion too. Results then are on higher side.


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4. Write any FOUR applications of COD.COD is used to measure:-

The Chemical Oxygen Demand, or COD,is a measurement of the amountof material that can be oxidized (combined with oxygen) in the presenceof a strong chemical oxidizing agent.

Indirectly measures the amount of organic compounds present in water. Determines the total amount of organic pollutants and oxygen

consumption is any water body like lakes or river and in the waste water

samples (Before and After treatments)

Indirect Idea of BOD.Since the COD test can be performed rapidly, it isoften used as a rough approximation of the water's BOD

5. Why COD values are always higher than BOD values?BOD contains only biodegradable while The COD test measures some additionalorganic matter (such as cellulose) which is not normally oxidized by biological

action. Thats why COD values are higher than BOD.

As with the BOD test, the COD test is reported as mg/Lit of oxygen used.

COMMENTS

COD value for Sample 1 is greater than for Sample 2 which showsthat Sample 1 contains more concentration of oxidizable matter i.e.

organic matter in it.

COD values suggest that BOD values will also be quite high eventhough always lesser than COD value.


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Kjeldhal NitrogenTitle

To determine the amount of nitrogen in given water sample.

GeneralAll nitrogen present in the organic compounds may be considered as organic nitrogen.

This includes amino acids, amines, amides, nitroderivatives and no of other organic

compounds. In waters and waste water the form of nitrogen of greatest intrest are organic

N and ammonia-N, nitrates and nitrites. Organic nitrogen is defined functionally asorganically bound nitrogen in tri negative state. In wastewaters organic-N include such

natural materials like proteins, peptides , nucleic acids, urea and numerous synthetic

organic materials. Organic nitrogen and ammonia can be determined together and have

been refered as kjeldhal nitrogen.

Principle:In the presence of H2SO4 , K2 S04 and CuSO4. Ammonia nitrogen of many organicmaterials is converted to ammonium sulfate. Free ammonia and ammonia nitrogen also

are converted to ammonium sulfate. During sample digestion a cuppric ammonia

complex is formed. After this mercury ammonia complex in the digestion has beendecomposed by sodium thiosulfate, the ammonia is distilled from an alkaline layer and

absorbed in boric acid.

Basic Steps involved:

1) Digestion2)

Distillation3) Titration.

Titration theory:The organic nitrogen is converted to ammonia nitrogen during the digestion. Boric acid isan excellent buffer. It combines with ammonia in the distillate to form ammonia and

borate ions.

NH3 + H3 BO3-- NH4++ H2BO3

-

The ammonia then is measured by back titration with strong acid such as sulfuric acid.

Actually the acid measures the amount of boric ion present in the solution as follows.H2BO3

-+H

+- H3BO3


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Reagents:

1) Digestion reagent ( dissolve 134g K2SO4 and 7.3g CuSO4 in about 800ml watercarefully add 134ml coc H2SO4. Cool to room temperature, dilute to 1 lit with

water . mix well keep at temp 200

C to prevent vaporization).2) Phenolphthalein indicator3) Sodium hydroxide4) Mixed indicator solution5) Indicating boric acid solution6) Standard sulfuric acid titrant.7) Hydroxide thiosulfate reagent.

Procedure:

Digestion

1. Take 280ml of sample in a kjeldhal flask.

2. Add few glass beads to it then add 50ml digestion reagent

3. Mix, heat and continue boiling until solution remains 25-50ml.

4. Cool it and add distilled water to it to make the volume 300ml.5. Add 0.5 ml phenolphthalein indicator.

6. Add 50 ml thiosulfate hydroxide reagent solution.

If pink color does not appear then add more 50ml thiosulfate hydroxide reagent solution.

Distillation

1. In collect the distillate in a flask containing boric acid solution.2. Collect 200ml distillate into 50ml boric acid solution.

Titration

1. Titrate it against 0.02N H2SO4 solution until color changes from purple to green.

2. Carry the blank titration, following all steps of procedure.


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Kjeldhal Nitrogen


Volume of Sample = 140 ml

A= volume of H2SO4 used for sample = 55 mlB= Volume of H2SO4 used for blank = 0.5 ml

Total nitrogen (mg/L)= 109 mg/l

Questions

1) What is Kjeldhal nitrogen?The total Kjeldhal nitrogen assay is an analytical chemistry method for the

quantitative determination of nitrogen in chemical substances, developed by JohnKjeldhal.

Organic nitrogen and ammonia determined together are referred as Kjeldhal

nitrogen.

2) What is blue baby disease?Nitrates also react directly with hemoglobin in human blood and other warm

blooded animals to produce methemoglobin. Methemoglobin destroys the ability

of red blood cells to transport oxygen. This condition is much severe in babies of

age 3 months. This is known as methemoglobinemia or blue baby syndrome.

3) How ammonia nitrogen can be determined?Ammonia nitrogen can be determined by several different methods. Selection ofspecific method must be based upon concentration level and amount and type of

interferences present.

In drinking water and some highly purified waste water effluent, it may bepossible to determine ammonia concentration directly by colorimetric methods.However, the approved method for waste water is preliminary distillation of the

Total Nitrogen (mg/L) = (A-B)* 280ml of sample


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ammonia into an acid adsorbing solution for colorimetric, titrimetric or specific

ion electrode determination.

4) Why hydroxide thiosulfate reagent is added in above experiment?Na2S2O3 is added to neutralize the residual chlorine in the sample.

5) Write the significance of nitrogen test in Environmental Engineering.Nitrogen is no doubt an essential component of all the living things. Excessiveconcentrations of certain nitrogen substances in some compartments of the

environment can lead to significant hazards and problems. So in order to deal with

these issues, first of all determination of concentration is vital.

COMMENTS

Value of total nitrogen is 109 mg/l.Organic nitrogen and ammonia combine together and give Kjeldhal

nitrogen.

More Kjeldhal nitrogen means more nitrogenous compounds arepresent in the sample.


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CHLORIDES

TITLE

Determination of chloride concentration in different water samples.

Chlorides

Chloride is a salt compound resulting from the combination of the gas chlorine and a

metal. Some common chlorides include sodium chloride (NaCl) and magnesium chloride

(MgCl2) and calcium chloride (CaCl2). Chlorine alone as Cl2 is highly toxic, and it isoften used as a disinfectant. In combination with a metal such as sodium it becomes

essential for life. Small amounts of chlorides are required for normal cell functions in

plant and animal life.

Chlorides in Water

Almost all natural waters contain chloride and sulfate ions. Their concentrations vary

considerably according to the mineral content of the earth in any given area. In smallamounts they are not significant. In large concentrations they present problems. Usually

chloride concentrations are low. Sulfates can be more troublesome because they generally

occur in greater concentrations. Low to moderate concentrations of both chloride andsulfate ions add palatability to water. In fact, they are desirable for this reason. Excessive

concentrations of either, of course, can make water unpleasant to drink.

The EPA Secondary Drinking Water Regulations recommend a maximum concentration

of 250 mg/lit for chloride ions and 250 mg/lit for sulfate ions (expressed as Cl-and S04

-2,

not as CaC03).

Chlorides give water a salty taste. At what concentrations this becomes noticeable againdepends upon the individual. In large concentrations chlorides cause a brackish, brinytaste that definitely is undesirable. Although chlorides are extremely soluble, they possess

marked stability. This enables them to resist change and to remain fairly constant in any

given water unless the supply is altered by dilution or by industrial or human wastes.Both chlorides and sulfates contribute to the total mineral content of water. As indicated

above, the total concentration of minerals may have a variety of effects in the home. High

concentrations of either sulfate or chloride ions add to the electrical conductivity of

water.Chlorides and sulfates can be substantially removed from water by reverse osmosis.

Deionization (demineralization) or distillation will also remove chlorides and sulfates

from water, but these methods are less suitable for household use than reverse osmosis.
http://www.freedrinkingwater.com/ro-45-detail.htmhttp://www.freedrinkingwater.com/ro-45-detail.htm


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Methods of determination of Chlorides

1. Mohrs method or Argentometric method2. Mercuric Nitrate method3.

Potentiometeric method

MERCURIC NITRATE METHOD

General Discussion

a. Principle:Chloride can be titrated with mercuric nitrate, Hg(N03)2 because of the formation of

soluble, slightly dissociated mercuric chloride. In the pH range 2.3 to 2.8, diphenyl car-

bazone indicates the titration end point by formation of a purple complex with the excess

mercuric ions. Xylene cyanol FF serves as a pH indicator and end-point enhancer.Increasing the strength of the titrant and modifying the indicator mixtures extend the

range of measurable chloride concentrations.

b. Interference:Bromide and iodide are titrated with Hg(N03)2 in the same manner as chloride. Chromate,

ferric, and sulfite ions interfere when present in excess of 10 mg/lit.

Procedure

A. for Titration of chloride concentrations less than 100 mg/lit:i. Take a 100-ml of chloride containing solution in titration flask (100ml chloride

containing sample is taken because of this sample contains less amount of

chlorides)

ii.

Add 1.0 ml indicator-acidifier reagent. It is a mixture of diphenyl carbazone andXylene cyanol. If

pH3.80 then the color of the solution will be blue

iii. Titrate this sample with 0.0141N Hg(N03)2 titrant till a definite purple end pointcolor achieved.

iv. We can use high strength 0.141N Hg(N03)2 instead of 0.0141N Hg(N03)2 as atitrant but after few drops the definite purple end point color will be achieved. So

it requires very care.v. Take three readings and determine mean volume.

vi. Take 100-ml distilled water in titration flask.vii. Add indicator (diphenyl carbazone + Xylene cyanol) + 10mg Na2C03.

viii. Titrate this sample with 0.0141N Hg(N03)2 titrant till a definite purple end pointcolor achieved.

ix. Take only reading and determine volume of the titrant Hg(N03)2 used


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x. Chloride ions concentration can be determined from following formula

100

10000x35.4xNx)BA(

mg/linCl

mlinsampleofvol

10000xClofEqWxNx)BA(mg/linCl

-

--

where:

A = mean volume of Hg(N03)2 in ml used to titrate the sampleB = mean volume of Hg(N03)2 in ml used to titrate the distilled water

N = normality of Hg(N03)2 i.e 0.0141N Hg(N03)2

B. for Titration of chloride concentrations greater than 100 mg/lit:i. Take a 50-ml of chloride containing solution in titration flask

ii. Add 0.5 ml mixed indicator. It is a mixture of diphenylcarbazone andbromphenol. If

iii. Add few drops of 0.1N HN03 till color become yellowiv. Titrate this sample with high strength 0.141N Hg(N03)2 titrant till a definite

purple end point color achievedv. Take three readings and determine mean volume.

vi. Take 50-ml distilled water in titration flask.vii. Add indicator

viii. Titrate this sample with 0.0141N Hg(N03)2 titrant till a definite purple end pointcolor achieved.

ix. Take only reading and determine volume of the titrant Hg(N03)2 usedx. Chloride ions concentration can be determined from following formula

50

10000x35.4xNx)BA(mg/linCl

mlinsampleofvol

10000xClofMWxNx)BA(mg/linCl

-

--

where:

A = mean volume of Hg(N03)2 in ml used to titrate the sampleB = mean volume of Hg(N03)2 in ml used to titrate the distilled water

N = normality of Hg(N03)2 i.e 0.10N Hg(N03)2


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CHLORIDES


Sampleno.

Initial

Volume

finalvolu

me

Differencechloride

concentration

(ml) (ml) (ml) (mg/l)

1 6 10.3 4.3 429.26

2 1 6.5 5.5 549.05

2 10.5 12.2 1.7 169.7

Questions

1) What is the WHO guideline value for chlorides?Health based guideline for chlorides in drinking water were proposed in 1993.

The EPA Secondary Drinking Water Regulations recommend a maximum

concentration of 250 mg/lit for chloride ions and 250 mg/lit for sulfate ions(expressed as Cl

-and S04

-2, not as CaC03).

2) What is the significance of chlorides test for construction purposes?Water soluble chlorides, in combination with oxygen and moisture can cause

corrosion of embedded metal in concrete. Unfortunately, chlorides are amongthe most abundant materials on the earth. Therefore, it is not practically possible

to have no chlorides present in the water of concrete mix. Due to the fact, ACI has

established some practical limits regarding this.

Usually upto 500 mg/lit concentration is acceptable.

3) How chlorides gain access to natural waters?Chlorides gain access to natural waters in many ways. Water has the ability todissolve chlorides from top soils and deep formations. Plus occurrence of

chlorides can be natural as well as anthropogenic. Use of inorganic fertilizers is

also a source of entering chlorides in water. Landfill leaches, septic tank effluents,animal feeds, industrial effluent, irrigation drainage and sea water intrusions are

coastal reasons.


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4) What would be the role of mixed indicator in this titration?Chloride can be titrated with mercuric nitrate, Hg(N03)2 because of the formationof soluble, slightly dissociated mercuric chloride. In the pH range 2.3 to 2.8,

diphenyl carbazone indicates the titration end point by formation of a purple

complex with the excess mercuric ions. Xylene cyanol FF serves as a pH indicatorand end-point enhancer. Increasing the strength of the titrant and modifying theindicator mixtures extend the range of measurable chloride concentrations.

5) Why different procedures are used depending upon chlorides concentration?The chloride content of natural surface waters will depend upon geology of the

area. In a limestone area like Richmand, natural surface waters will have very

little chlorides in it. (10-50) mg/l . Appreciable higher chloride concentrationswould suggest contaminations. Concentration of chlorides can be damaging to

metal pipes as well as agricultural crops.

Different procedures are used depending upon concentration is due to the factthat chloride ions with Hg+2

to form HgCl2 (in concentrations greater than 100

mg/l) which white ppts are insoluble and will faint the color complex and end

point can not be indicated properly.

COMMENTS

The chloride ion concentration for sample 1 and sample 2 is 429.26mg/l and 549.05 mg/l respectively. Which is greater than 250 mg/l.

since 250 mg/l is a threashold value given by WHO guidelines so bothsample 1 and 2 are rejected.

Sample 3 is suitable because Cl- concentration is less than 250 mg/l.Sample 2 and sample 3 may be used for concrete mix because their

concentration is less than 500 mg/l.

Documents

3-Env. Engg.-II Lab Civ-10-36